Points of Inflection Exercises

Example 1

For the function, find all points of inflection or determine that no such points exist.

f (x) = sin x

Answer

f (x) = sin x

The first derivative is f '(x) = cos x and the second derivative is

f "(x) = -sin x.

The second derivative is never undefined, and is zero at nπ for every integer n. Therefore our inflection points are nπ for every integer n. Ifwe look at the graph, this makes sense. We can think of the graph of sin x as a series of upside down and right-side up bowls, and an inflection point occurs between every two bowls:

Example 2

For the function, find all points of inflection or determine that no such points exist.

f (x) = ex

Answer

f (x) = ex

Every time we take the derivative of ex we find ex again, therefore the first and second (and all other) derivatives of f (x) are ex. Since f"(x) = ex is never undefined and is never zero, the function f (x) = ex has no inflection points. This makes sense, since the graph of f (x) = ex is concave up everywhere:

Example 3

For the function, find all points of inflection or determine that no such points exist.

Answer

We need to use some quotient rule stuff here. First, the first derivative:

Usually we say not to bother squaring the denominator when we use the quotient rule. However, this denominator is only one term, so it looks better after we square it.

The second derivative has no roots and is undefined at x = 0. However, since x = 0 isn't in the domain of f to begin with, it can't be an inflection point. Therefore the function f (x) has no inflection points, which indeed it does not, it's concave up everywhere it is defined:

This might seem like a lot of work for no inflection points. However, knowing a graph has no inflection points is still a useful piece of information.

Example 4

For the function, find all points of inflection or determine that no such points exist.

Answer

Since f " is a polynomial it is never undefined. Its roots are x = 2 and x = 3:

When x < 2, both x - 2 and x - 3 are negative, so f " is positive:

When x is between 2 and 3, the quantity x - 2 is positive but x - 3 is negative so f " is negative:

Finally, when x is greater than 3 both x - 2 and x - 3 are positive, so f " is positive:

Since the sign of f " changes at x = 2 and x = 3, these are both inflection points.

Example 5

For the function, find all points of inflection or determine that no such points exist.

f (x) = 5x + 2

Answer

f (x) = 5x + 2

The derivative calculations for this one are super-quick (f '(x) = 5 and f "(x) = 0), but you don't even need to do that. Look at the graph:

A line has no concavity anywhere. Therefore it never changes concavity, so it can't have any inflection points. The end!

Example 6

For the function, find all points of inflection or determine that no such points exist.

f (x) = x1/3

Answer

f (x) = x1/3

We find the first two derivatives:

This function is undefined when x = 0. Since x = 0 is in the domain of f, this is a possible inflection point. To determine if x = 0 is a real inflection point, we need to inspect the sign of f " to either side.

When x is greater than zero,

is also greater than zero, so

is negative.

When x is less than zero,

is also less than zero because of the odd numbers in the fractional exponent (the cube root of a negative number is negative, and raising a negative number to the 5th power is negative). Therefore

is positive.

Since f " changes sign at x = 0, this is an inflection point. Here's the graph:

Example 7

For the function, find all points of inflection or determine that no such points exist.

f (x) = xex

Answer

f (x) = xex

We find the derivatives using the product rule:

This is never undefined, and has its only root at x = -2. Since ex is always positive, it doesn't affect the sign of f". The only part that affects the sign of f" is the (2 + x):

f"(x) = ex(2 + x)

Since 2 + x is negative for x < -2 and positive for x > -2, the sign of f" changes at x = -2, so this is an inflection point. Here's the graph:

Example 8

For the function, find all points of inflection or determine that no such points exist.

Answer

We use the quotient rule a couple of times to find the first two derivatives:

The second derivative is undefined when x = 0, but that's not in the domain of f anyway so it doesn't count. The second derivative is 0 when

To confirm that this is an inflection point, we need to check what the sign of f" does as we cross x = e3/2 ≅ 4.48.

The numbers x = 4 and x = 5 are nice ones to use.

Since the sign of f" does change, x = e3/2 is indeed an IP.

From the graph, this is believable:

Example 9

For the function, find all points of inflection or determine that no such points exist.

f (x) = x lnx for x > > 0

Answer

f (x) = x lnx for x > 0

The first derivative is

and the second derivative is

The second derivative is never zero, and is undefined only for x = 0. However, x = 0 isn't in the domain of the function anyway, so it can't be an IP. Thisfunction has no inflection points, which makes sense because it appears to be concave up the whole way:

Example 10

For the function, find all points of inflection or determine that no such points exist.

The logistic equation

Hint

Rewrite the function before taking derivatives. Oh, and be patient.

Answer

We don't need to use the quotient rule. First, rewrite the function:

f (x) = (1 + 5e-x)-1

Now use the chain rule to take the derivative:

To take the second derivative we need to use the product rule and the chain rule:

This is the second derivative. But it's ugly. Can we make it more clear? Since -2 = -3 + 1, we can replace

(1 + 5e-x)-2

with

(1 + 5e-x)-3(1 + 5e-x).

Believe us? Good. This is what we have now:

Phew! This looks better, since it's all broken down into factors.

f" can never be undefined. Is it ever zero? Go through one factor at a time.

f"(x) = -5e-x (1 + 5e-x)-3[-5e-x + 1]

The factor 5e-x is never zero.

f"(x) = -5e-x(1 + 5e-x)-3[-5e-x + 1]

Since 5e-x is always positive, so is (1 + 5e-x), and so is (1 + 5e-x)-3. That part is never zero.

f"(x) = -5e-x (1 + 5e-x)-3[-5e-x + 1]

This part can equal zero if

0 = -5e-x + 1.

Solve:

Since ln e= 1, therefore

Remember, this is still only a possible IP. Plug in numbers to the left and right of our possible IP, and see what happens to the sign of f". We already know the only factor of f" that isn't always positive is the last one:

Therefore if the sign of [-5e-x + 1] changes from one side to the other of our possible IP, the sign of f" changes also, and so we really do have an inflection point.

For a point to the left we'll use x = 1, and for a point to the right use x = 2. Pull out the calculator:

The sign of [-5e-x + 1] changes from one side to the other of our possible IP, so we really do have an IP at